1. Given position coordinates

The particle has coordinates: $x=a\cos \omega t, \qquad y=a\sin \omega t, \qquad z=a\omega t$

We need the speed, i.e. the magnitude of velocity: $v=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}$

2. Differentiate each coordinate w.r.t. time

$\frac{dx}{dt}=-a\omega \sin \omega t$

$\frac{dy}{dt}=a\omega \cos \omega t$

$\frac{dz}{dt}=a\omega$

3. Compute speed

Substitute into the velocity magnitude formula:

$v=\sqrt{(-a\omega \sin \omega t)^2+(a\omega \cos \omega t)^2+(a\omega)^2}$

$v=\sqrt{a^2\omega^2\sin^2\omega t+a^2\omega^2\cos^2\omega t+a^2\omega^2}$

Factor out $a^2\omega^2$:

$v=\sqrt{a^2\omega^2(\sin^2\omega t+\cos^2\omega t+1)}$

Using $\sin^2\theta+\cos^2\theta=1$ we get:

$v=\sqrt{a^2\omega^2(1+1)}$

$v=\sqrt{2a^2\omega^2}=\sqrt{2}\,a\omega$

4. Match with options

The correct option is:

$\boxed{A\;:\;\sqrt{2}\,a\omega}$